Problem: $f(x)=\begin{cases} 2^x&\text{for }0<x\leq4 \\\\ 8\sqrt{x}&\text{for }x>4 \end{cases}$ Find $\lim_{x\to 4}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $8$ (Choice C) C $16$ (Choice D) D The limit doesn't exist.
$x=4$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to 4}f(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $4$ from the left. We will use the fact that $f(x)=2^x$ for $x$ -values smaller than $4$. $\begin{aligned} &\phantom{=}\lim_{x\to 4^-}f(x) \\\\ &=\lim_{x\to 4^-}2^x \\\\ &=2^4&\gray{\text{Direct substitution}} \\\\ &=16 \end{aligned}$ Let's find the limit as $x$ approaches $4$ from the right. We will use the fact that $f(x)=8\sqrt{x}$ for $x$ -values greater than $4$. $\begin{aligned} &\phantom{=}\lim_{x\to 4^+}f(x) \\\\ &=\lim_{x\to 4^+}8\sqrt{x} \\\\ &=8\sqrt{4}&\gray{\text{Direct substitution}} \\\\ &=16 \end{aligned}$ The one-sided limits are both equal to $16$. This means that $\lim_{x\to 4}f(x)=16$.